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QJEEAdv18Ch2267: JEE Advanced 2018 Chemistry - Class 12 JEEAdv18Ch2267 Solution: LIST-I contains reactions and LIST-II contains major products.…

LIST-I contains reactions and LIST-II contains major products.

JEE Advanced 2018 Paper 2 Offline Chemistry - Alcohols, Phenols and Ethers Question 21 English

Match the reaction in LIST-I with one or more products in LIST-II and choose the correct option.

Multiple Choice Options

  • A P−1,5;Q−2;R−3;S−4
  • B P−1,4;Q−2;R−4;S−3
  • C P−1,4;Q−1,2;R−3,4;S−4
  • D P−4,5;Q−4;R−4;S−3,4

Step-by-Step Verified Solution

P. Reactant 1 , acts as a base and abstracts hydrogen from β\beta carbon of tert butyl bromide halide (reactant 2 ) forming an alkene and an alcohol.

JEE Advanced 2018 Paper 2 Offline Chemistry - Alcohols, Phenols and Ethers Question 21 English Explanation 1
Elimination proceeds via E2\mathrm{E}_2 mechanism. Substitution of reactant 1 is not possible on reactant 2 due to steric crowding at 2.

Q. Reactant 1 (tert butoxide) acts as a base and abstracts proton from acid (HBr)(\mathrm{HBr}) forming an alcohol and methyl bromide.

JEE Advanced 2018 Paper 2 Offline Chemistry - Alcohols, Phenols and Ethers Question 21 English Explanation 2
R. Sodium methoxide ( NaOMe)\mathrm{NaOMe}) is a strong base and abstracts proton from β\beta carbon of tert Butyl with simultaneous less of bromide ion forming an alkene.

JEE Advanced 2018 Paper 2 Offline Chemistry - Alcohols, Phenols and Ethers Question 21 English Explanation 3
S. Reactant 1 (tert-butoxide) acts as a nucleophile and attacks from the backside of methyl bromide, with subsequent removal of bromide ion. This is an SN2\mathrm{S}_{\mathrm{N}} 2 mechanism.

JEE Advanced 2018 Paper 2 Offline Chemistry - Alcohols, Phenols and Ethers Question 21 English Explanation 4

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